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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2017 · Mathematics · Quadratic Equation

If α and β are the roots of the equation 3 x^2+2 x+1=0 , then the...

If α\alpha and β\beta are the roots of the equation 3x2+2x+1=03 x^2+2 x+1=0, then the equation whose roots are α+β−1\alpha+\beta^{-1} and β+α−1\beta+\alpha^{-1} is

Options

  1. A3x2+8x+16=03 x^2+8 x+16=0
  2. B3x2−8x−16=03 x^2-8 x-16=0
  3. C3x2+8x−16=03 x^2+8 x-16=0
  4. Dx2+8x+16=0x^2+8 x+16=0

Correct answer

A. 3x2+8x+16=03 x^2+8 x+16=0

Step-by-step solution

3x2+2x+1=03 x^2+2 x+1=0 Sum of the roots =α+β=−23=\alpha+\beta=\frac{-2}{3}\ldots(1) Productoftheroots=α⋅β=13of the roots=\alpha \cdot \beta=\frac{1}{3}\ldots(2) We havetofindtheequationwiththerootsα+β−1andβ+α−1to find the equation with the roots\alpha+\beta^{-1} and \beta+\alpha^{-1}. Sum of the roots (S)=α+β−1+β+α−1=α+1β+β+1α(S)=\alpha+\beta^{-1}+\beta+\alpha^{-1}=\alpha+\frac{1}{\beta}+\beta+\frac{1}{\alpha}=(α+β)+(α+βαβ)=−23+(−2313)( from  (1),(2)) \begin{aligned} & =(\alpha+\beta)+\left(\frac{\alpha+\beta}{\alpha \beta}\right) \\ & =\frac{-2}{3}+\left(\frac{\frac{-2}{3}}{\frac{1}{3}}\right) \quad(\text{\text{ from } (1),(2)) }\end{aligned}=−23−2=−2−63=−83.=\frac{-2}{3}-2=\frac{-2-6}{3}=\frac{-8}{3} . Product of the roots (P)=(α+β−1)(β+α−1)=(α+1β)(β+1α)=αβ+1+1+1αβ=αβ+2+1αβ=13+2+113=13+2+3=13+5=163(\mathrm{P})=\left(\alpha+\beta^{-1}\right)\left(\beta+\alpha^{-1}\right)=\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)=\alpha \beta+1+1+\frac{1}{\alpha \beta}=\alpha \beta+2+\frac{1}{\alpha \beta}=\frac{1}{3}+2+\frac{1}{\frac{1}{3}}=\frac{1}{3}+2+3=\frac{1}{3}+5=\frac{16}{3}. So, the required equation is x2−S.x.+P=0=x2+83x+163=0⇒3x2+8x+16=0x^2-S . x .+P=0=x^2+\frac{8}{3} x+\frac{16}{3}=0 \Rightarrow 3 x^2+8 x+16=0
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